Let f: R to (0, infty)$f: R \to (0, \infty)$ be a twice differentiable function such that f(3) = 18$f(3) = 18$ , f'(3) = 0$f'(3) = 0$ and f''(3) = 4$f''(3) = 4$ .
Then lim_x to 1 left( log_e left( fracf(2 + x)f(3) right)^frac18(x-1)^2 right)$\lim_{x \to 1} \left( \log_{e} \left( \frac{f(2 + x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$ is equal to:
A.1
B.9
C.2
D.18
Solution & Explanation
### Related Formula
For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals:
e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$
### Core Logic
Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$.
As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$.
This is a standard 1^infty$1^{\infty}$ form.
T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$
### Step 1: Simplify Exponent Limit
Given f(3) = 18$f(3) = 18$:
T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$
This is a frac00$\frac{0}{0}$ form limit.
### Step 2: Apply L'Hopital's Rule
Differentiate numerator and denominator w.r.t x$x$:
T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$
This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$.
Apply L'Hopital's Rule again:
T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$
Substitute x = 1$x = 1$:
T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$
### Step 3: Final Calculation
Given f''(3) = 4$f''(3) = 4$:
T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$
The question asks for log_e(T)$\log_e(T)$:
log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$
### Pattern Recognition
When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Limits and Derivatives
Class 12 Maths: Limits, Continuity and Differentiability
Keywords:#twice differentiable function#JEE Main 2026 Morning Q10#Limits, Continuity and Differentiability JEE Main 2026#1^infinity Limit Form with L'Hopital's Rule JEE Main 2026
More Limits, Continuity and Differentiability Previous-Year Questions — Page 5
Q68jee_main_2025_24_jan_eveningNumber of Real Solutions of Equations
The number of real solution(s) of the equation x^2+3x+2=min\|x-3|, |x+2|\$x^{2}+3x+2=\min\{|x-3|, |x+2|\}$ is: [cite: 3396, 3397]
A.2$2$
B.0$0$
C.3$3$
D.1$1$
Solution
### Related Formula
The function min\f(x), g(x)\$\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate x$x$.
### Core Logic
Analyze the conditions for the \right-hand function min\|x-3|, |x+2|\$\min\{|x-3|, |x+2|\}$:
- The intersection of |x-3| = |x+2|$|x-3| = |x+2|$ happens at x - 3 = -(x + 2) Rightarrow 2x = 1 Rightarrow x = 0.5$x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$.
- For x le 0.5$x \le 0.5$, |x+2| le |x-3| Rightarrow min = |x+2|$|x+2| \le |x-3| \Rightarrow \min = |x+2|$.
- For x > 0.5$x > 0.5$, |x-3| le |x+2| Rightarrow min = |x-3|$|x-3| \le |x+2| \Rightarrow \min = |x-3|$.
Min function intersection graph for Q68 - JEE Main 2025 Evening
### Step 1: Check Interval x le -2$x \le -2$
Here, |x+2| = -(x+2) = -x-2$|x+2| = -(x+2) = -x-2$:
x^2 + 3x + 2 = -x - 2 Rightarrow x^2 + 4x + 4 = 0$$x^2 + 3x + 2 = -x - 2 \Rightarrow x^2 + 4x + 4 = 0$$(x+2)^2 = 0 Rightarrow x = -2$$(x+2)^2 = 0 \Rightarrow x = -2$$
This is a valid solution as it lies precisely within the interval condition boundary.
### Step 2: Check Interval -2 < x le 0.5$-2 < x \le 0.5$
Here, |x+2| = x+2$|x+2| = x+2$:
x^2 + 3x + 2 = x + 2 Rightarrow x^2 + 2x = 0$$x^2 + 3x + 2 = x + 2 \Rightarrow x^2 + 2x = 0$$x(x+2) = 0 Rightarrow x = 0 quad textor quad x = -2$$x(x+2) = 0 \Rightarrow x = 0 \quad \text{or} \quad x = -2$$
Only x = 0$x = 0$ fits inside this interval.
### Step 3: Check Interval x > 0.5$x > 0.5$
Here, min = |x-3| = 3-x$\min = |x-3| = 3-x$:
x^2 + 3x + 2 = 3 - x Rightarrow x^2 + 4x - 1 = 0$$x^2 + 3x + 2 = 3 - x \Rightarrow x^2 + 4x - 1 = 0$$x = frac-4 pm sqrt16 - 4(1)(-1)2 = -2 pm sqrt5$$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$$
Evaluating values: -2 + sqrt5 approx 0.236$-2 + \sqrt{5} \approx 0.236$, which does not satisfy x > 0.5$x > 0.5$. Thus, no real roots occur in this span.
Combining valid points, we find exactly 2 distinct real solutions (x = -2, 0$x = -2, 0$).
### Pattern Recognition
Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Quadratic Equations
Class 12 Mathematics: Limits, Continuity and Differentiability
Q57jee_main_2025_24_jan_morningLimits of Algebraic and Trigonometric Functions
lim_x to 0 csc x left(sqrt2cos^2 x + 3cos x - sqrtcos^2 x + sin x + 4right)$\lim_{x \to 0} \csc x \left(\sqrt{2\cos^2 x + 3\cos x} - \sqrt{\cos^2 x + \sin x + 4}\right)$ is equal to :
A.0$0$
B.frac12sqrt5$\frac{1}{2\sqrt{5}}$
C.frac1sqrt15$\frac{1}{\sqrt{15}}$
D.-frac12sqrt5$-\frac{1}{2\sqrt{5}}$
Solution
### Related Formula
To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching:
(sqrtA - sqrtB)(sqrtA + sqrtB) = A - B$$(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B$$
### Core Logic
Rewrite the expression as a fraction with sin x$\sin x$ in the denominator and rationalize the numerator:
lim_x to 0 frac(2cos^2 x + 3cos x) - (cos^2 x + sin x + 4)sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$$\lim_{x \to 0} \frac{(2\cos^2 x + 3\cos x) - (\cos^2 x + \sin x + 4)}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$$= lim_x to 0 fraccos^2 x + 3cos x - sin x - 4sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$$= \lim_{x \to 0} \frac{\cos^2 x + 3\cos x - \sin x - 4}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$$
### Step 1: Simplify Numerator and Group Terms
Express the numerator terms to isolate algebraic patterns:
cos^2 x + 3cos x - 4 - sin x = (cos x - 1)(cos x + 4) - sin x$$\cos^2 x + 3\cos x - 4 - \sin x = (\cos x - 1)(\cos x + 4) - \sin x$$
Substitute this back into our rationalized limit format:
= lim_x to 0 frac(cos x - 1)(cos x + 4) - sin xsin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$$= \lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$$
### Step 2: Distribute sin x$\sin x$ in Denominator
Split the limit across the two separated numerator expressions:
= lim_x to 0 left[ fraccos x - 1sin x cdot (cos x + 4) - 1 right] cdot frac1sqrt2(1)+3 + sqrt1+0+4$$= \lim_{x \to 0} \left[ \frac{\cos x - 1}{\sin x} \cdot (\cos x + 4) - 1 \right] \cdot \frac{1}{\sqrt{2(1)+3} + \sqrt{1+0+4}}$$
Evaluate the limit component values:
lim_x to 0 fraccos x - 1sin x = lim_x to 0 frac-2sin^2(x/2)2sin(x/2)cos(x/2) = lim_x to 0 [-tan(x/2)] = 0$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = \lim_{x \to 0} \frac{-2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} [-\tan(x/2)] = 0$$
Substituting this zero value simplifies the numerator expression directly:
= left[ 0 cdot (1 + 4) - 1 right] cdot frac1sqrt5 + sqrt5 = frac-12sqrt5$$= \left[ 0 \cdot (1 + 4) - 1 \right] \cdot \frac{1}{\sqrt{5} + \sqrt{5}} = \frac{-1}{2\sqrt{5}}$$
### Pattern Recognition
Recognizing that fraccos x - 1sin x to 0$\frac{\cos x - 1}{\sin x} \to 0$ as x to 0$x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Limits and Derivatives
Q72jee_main_2025_28_jan_eveningLimits of Trigonometric Functions
Let f(x)=lim_nrightarrow inftysum_r=0^nleft(fractan(x/2^r+1)+tan^3(x/2^r+1)1-tan^2(x/2^r+1)right).$f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1})+\tan^{3}(x/2^{r+1})}{1-\tan^{2}(x/2^{r+1})}\right).$ Then lim_xrightarrow0frace^x-e^f(x)(x-f(x))$\lim_{x\rightarrow0}\frac{e^{x}-e^{f(x)}}{(x-f(x))}$ is equal to
Numerical Answer.Answer: 1 to 1
Solution
### Related Formula
Trigonometric identity:
fractantheta + tan^3theta1-tan^2theta = tantheta left( frac1+tan^2theta1-tan^2theta right) = fractanthetacos 2theta$$\frac{\tan\theta + \tan^3\theta}{1-\tan^2\theta} = \tan\theta \left( \frac{1+\tan^2\theta}{1-\tan^2\theta} \right) = \frac{\tan\theta}{\cos 2\theta}$$
Also note standard telescopic identity:
tan 2phi - tanphi = fractanphicos 2phi$$\tan 2\phi - \tan\phi = \frac{\tan\phi}{\cos 2\phi}$$
### Core Logic
Let theta = fracx2^r+1$\theta = \frac{x}{2^{r+1}}$. The term inside the summation simplifies to:
tanleft(fracx2^rright) - tanleft(fracx2^r+1right)$$\tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$$
Now, evaluating the summation:
sum_r=0^n left[ tanleft(fracx2^rright) - tanleft(fracx2^r+1right) right] = tan x - tanleft(fracx2^n+1right)$$\sum_{r=0}^{n} \left[ \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right) \right] = \tan x - \tan\left(\frac{x}{2^{n+1}}\right)$$
Taking the limit as n to infty$n \to \infty$, tanleft(fracx2^n+1right) to tan(0) = 0$\tan\left(\frac{x}{2^{n+1}}\right) \to \tan(0) = 0$.
Therefore, f(x) = tan x$f(x) = \tan x$.
### Step 1: Evaluate the Limit
We need to find:
lim_xrightarrow0frace^x-e^tan xx-tan x$$\lim_{x\rightarrow0}\frac{e^{x}-e^{\tan x}}{x-\tan x}$$
Factor out e^tan x$e^{\tan x}$ from the numerator:
lim_xrightarrow0 e^tan x cdot left[ frace^x-tan x - 1x-tan x right]$$\lim_{x\rightarrow0} e^{\tan x} \cdot \left[ \frac{e^{x-\tan x} - 1}{x-\tan x} \right]$$
Let u = x - tan x$u = x - \tan x$. As x to 0$x \to 0$, u to 0$u \to 0$. The limit becomes:
lim_urightarrow0 e^0 cdot left[ frace^u - 1u right] = 1 times 1 = 1$$\lim_{u\rightarrow0} e^0 \cdot \left[ \frac{e^u - 1}{u} \right] = 1 \times 1 = 1$$
### Pattern Recognition
Standard limit substitution lim_y to 0 frace^y - 1y = 1$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ applies cleanly whenever the argument in the exponent matches the entire denominator layout.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometry
Class 12 Mathematics: Limits, Continuity and Differentiability
Qjee_main_2025_29_jan_morningLimits of Special Series
The value of lim_nto inftyleft(sum_K = 1^nfrack^3 + 6k^2 + 11k + 5(k + 3)!right)$\lim_{n\to \infty}\left(\sum_{K = 1}^{n}\frac{k^3 + 6k^2 + 11k + 5}{(k + 3)!}\right)$ is:
A.frac43$\frac{4}{3}$
B. 2
C.frac73$\frac{7}{3}$
D.frac53$\frac{5}{3}$
Solution
### Related Formula
sum_k=1^infty left( frac1k! - frac1(k+3)! right) implies textTelescoping Series simplification$$\sum_{k=1}^{\infty} \left( \frac{1}{k!} - \frac{1}{(k+3)!} \right) \implies \text{Telescoping Series simplification}$$
### Core Logic
Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base (k+3)$(k+3)$:
k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$$k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$$
### Step 1: Simplify General Term
T_k = frac(k+1)(k+2)(k+3)(k+3)! - frac1(k+3)!$$T_k = \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!}$$T_k = frac1k! - frac1(k+3)!$$T_k = \frac{1}{k!} - \frac{1}{(k+3)!}$$
This creates a clean telescoping layout format structure.
### Step 2: Sum the Series
Writing out expanded \partial sums up to infinity:
S = left( frac11! + frac12! + frac13! + frac14! + dots right) - left( frac14! + frac15! + frac16! + dots right)$$S = \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \right) - \left( \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \dots \right)$$
All higher terms cancel out systematically, leaving exactly the leading remaining fragments:
S = frac11! + frac12! + frac13! = 1 + frac12 + frac16 = frac106 = frac53$$S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} = 1 + \frac{1}{2} + \frac{1}{6} = \frac{10}{6} = \frac{5}{3}$$
### Pattern Recognition
Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums (V_n - V_n-k$V_n - V_{n-k}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Limits, Continuity and Differentiability
Class 11 Mathematics: Sequences and Series
Q73jee_main_2025_29_jan_morningSandwich Theorem and Greatest Integer Function
Let [t] be the greatest integer less than or equal to t. Then the least value of mathbfp in mathbbN$\mathbf{p} \in \mathbb{N}$ for which lim_mathrmx to 0^+ left( mathrmx left( left[ frac1mathrmx right] + left[ frac2mathrmx right] + dots + left[ fracmathrmpmathrmx right] right) - mathrmx^2 left( left[ frac1mathrmx^2 right] + left[ frac2^2mathrmx^2 right] + dots + left[ frac9^2mathrmx^2 right] right) right) geq 1$\lim_{\mathrm{x} \to 0^{+}} \left( \mathrm{x} \left( \left[ \frac{1}{\mathrm{x}} \right] + \left[ \frac{2}{\mathrm{x}} \right] + \dots + \left[ \frac{\mathrm{p}}{\mathrm{x}} \right] \right) - \mathrm{x}^{2} \left( \left[ \frac{1}{\mathrm{x}^{2}} \right] + \left[ \frac{2^{2}}{\mathrm{x}^{2}} \right] + \dots + \left[ \frac{9^{2}}{\mathrm{x}^{2}} \right] \right) \right) \geq 1$ is equal to
Numerical Answer.Answer: 24
Solution
### Related Formula
lim_x to 0^+ x left[ frackx right] = k$$\lim_{x \to 0^+} x \left[ \frac{k}{x} \right] = k$$sum_k=1^n k = fracn(n+1)2, quad sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$
### Core Logic
Using properties of Greatest Integer Function limits, as x to 0^+$x \to 0^+$, fraction values diverge cleanly to continuous variable distributions:
lim_x to 0^+ x left[ frackx right] = k implies sum_k=1^p k = fracp(p+1)2$$\lim_{x \to 0^+} x \left[ \frac{k}{x} \right] = k \implies \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$$
Similarly, for the second block component part:
lim_x to 0^+ x^2 left[ frack^2x^2 right] = k^2 implies sum_k=1^9 k^2 = frac9 times 10 times 196 = 285$$\lim_{x \to 0^+} x^2 \left[ \frac{k^2}{x^2} \right] = k^2 \implies \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285$$
### Step 1: Setup Inequality Formulation
Combine evaluated limits component parts:
fracp(p+1)2 - 285 geq 1$$\frac{p(p+1)}{2} - 285 \geq 1$$fracp(p+1)2 geq 286 implies p(p+1) geq 572$$\frac{p(p+1)}{2} \geq 286 \implies p(p+1) \geq 572$$
### Step 2: Solve for least natural number
Evaluate product bounds of adjacent integers:
If p = 23 implies 23 times 24 = 552$p = 23 \implies 23 \times 24 = 552$ (False)
If p = 24 implies 24 times 25 = 600$p = 24 \implies 24 \times 25 = 600$ (True)
Therefore, the least natural value of p$p$ is 24$24$.
### Pattern Recognition
Greatest Integer fractions simplify directly to standard scalar values inside limits evaluated at infinity or zero, letting you drop brackets and treat them as arithmetic sequences.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Limits
More Limits, Continuity and Differentiability Questions — jee_main_2026_21_jan_morning
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